Done with her new problems, Wendy takes a break from math. Still without any fresh reading material, she feels a bit antsy. She starts to feel annoyed that Michael's loose papers clutter the family van. Several of them are ripped, and bits of paper litter the floor. Tired of trying to get Michael to clean up after himself, Wendy spends a couple of minutes putting Michael's loose papers in the trash. "That seems fair to me," confirms Hannah encouragingly.
While collecting Michael's scraps, Wendy comes across a corner of a piece of paper with part of a math problem written on it. There is a monic polynomial of degree $n$, with real coefficients. The first two terms after $x^n$ are $a_{n-1}x^{n-1}$ and $a_{n-2}x^{n-2}$, but the rest of the polynomial is cut off where Michael's page is ripped. Wendy barely makes out a little of Michael's scribbling, showing that $a_{n-1}=-a_{n-2}$. Wendy deciphers the goal of the problem, which is to find the sum of the squares of the roots of the polynomial. Wendy knows neither the value of $n$, nor the value of $a_{n-1}$, but still she finds a [greatest] lower bound for the answer to the problem. Find the absolute value of that lower bound.

By Vieta's Formulas, given that $r_1, r_2, \cdots r_n$ are roots of the polynomial, we know that $\sum_{i=1}^n r_i = -a_{n-1}$ and $r_1r_2 + r_1r_3 \cdots r_{n-1}r_n = a_{n-2}$.
From the equation $\sum_{i=1}^n r_i = -a_{n-1}$, squaring both sides and substituting results in\begin{align*} \sum_{i=1}^n r_i^2 + 2(r_1r_2 + r_1r_3 \cdots r_{n-1}r_n) &= (a_{n-1})^2 \\ \sum_{i=1}^n r_i^2 + 2a_{n-2} &= (-a_{n-2})^2 \\ \sum_{i=1}^n r_i^2 &= (a_{n-2})^2 - 2a_{n-2} \end{align*}To find the lower bound of $\sum_{i=1}^n r_i^2$, we need to find the lower bound of $(a_{n-2})^2 - 2a_{n-2}$. The minimum of the quadratic is $-1$, so the absolute value of the lower bound of the sum of the squares is $\boxed{1}$.